We need to solve problems 9, 10, 11 and 12, finding the discriminant and stating the number and type of solutions for each quadratic equation. We also need to factor completely problems 13, 14, 15 and 16.

1. Problem Description

We need to solve problems 9, 10, 11 and 12, finding the discriminant and stating the number and type of solutions for each quadratic equation. We also need to factor completely problems 13, 14, 15 and

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2. Solution Steps

Problem 9: $-4n^2 + 4n - 1 = 0$

$a = -4, b = 4, c = -1$

Discriminant = $b^2 - 4ac = 4^2 - 4(-4)(-1) = 16 - 16 = 0$

Since the discriminant is 0, there is 1 real solution.

Problem 10: $-5x^2 - x + 9 = 9$

$-5x^2 - x = 0$

$a = -5, b = -1, c = 0$

Discriminant = $b^2 - 4ac = (-1)^2 - 4(-5)(0) = 1 - 0 = 1$

Since the discriminant is positive, there are 2 real solutions.

Problem 11: $-m^2 + 2m = 1$

$-m^2 + 2m - 1 = 0$

$a = -1, b = 2, c = -1$

Discriminant = $b^2 - 4ac = 2^2 - 4(-1)(-1) = 4 - 4 = 0$

Since the discriminant is 0, there is 1 real solution.

Problem 12: $2x^2 - 5 = 7x^2 - 5 - 9x$

$0 = 5x^2 - 9x$

$a = 5, b = -9, c = 0$

Discriminant = $b^2 - 4ac = (-9)^2 - 4(5)(0) = 81 - 0 = 81$

Since the discriminant is positive, there are 2 real solutions.

Problem 13: $4v^2 - 8v - 140$

First, factor out the greatest common factor (GCF), which is 4:

$4(v^2 - 2v - 35)$

Now, factor the quadratic expression inside the parentheses:

$v^2 - 2v - 35 = (v - 7)(v + 5)$

So the complete factorization is $4(v - 7)(v + 5)$

Problem 14: $9r^3 - 3r^2 - 72r$

First, factor out the greatest common factor (GCF), which is $3r$:

$3r(3r^2 - r - 24)$

Now, factor the quadratic expression inside the parentheses:

$3r^2 - r - 24$

We look for two numbers that multiply to $3*(-24) = -72$ and add up to $-1$. These numbers are $-9$ and $8$.

So, we can rewrite the middle term as $-9r + 8r$:

$3r^2 - 9r + 8r - 24$

Now, factor by grouping:

$3r(r - 3) + 8(r - 3)$

$(3r + 8)(r - 3)$

So the complete factorization is $3r(3r + 8)(r - 3)$

Problem 15: $48x^2 + 168x$

First, factor out the greatest common factor (GCF). The GCF of 48 and 168 is 24, and both terms have an $x$, so the GCF is $24x$.

$24x(2x + 7)$

Problem 16: $30x^3 - 309x^2 + 90x$

First, factor out the greatest common factor (GCF). The GCF of 30, 309, and 90 is 3, and all terms have an $x$, so the GCF is $3x$.

$3x(10x^2 - 103x + 30)$

Now, we need to factor the quadratic $10x^2 - 103x + 30$.

We're looking for two numbers that multiply to $10*30 = 300$ and add up to $-103$.

These numbers are $-100$ and $-3$.

$10x^2 - 100x - 3x + 30$

$10x(x - 10) - 3(x - 10)$

$(10x - 3)(x - 10)$

Thus, the complete factorization is $3x(10x - 3)(x - 10)$

3. Final Answer

Problem 9:

Discriminant = 0, 1 real solution

Problem 10:

Discriminant = 1, 2 real solutions

Problem 11:

Discriminant = 0, 1 real solution

Problem 12:

Discriminant = 81, 2 real solutions

Problem 13:

$4(v - 7)(v + 5)$

Problem 14:

$3r(3r + 8)(r - 3)$

Problem 15:

$24x(2x + 7)$

Problem 16:

$3x(10x - 3)(x - 10)$